Degree Of Unsaturation

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A hydrocarbon with the molecular formula $C_6H_{10}$ is being analysed, and the total number of degrees of unsaturation present in the molecule must be determined.

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The degree of unsaturation, also called the index of hydrogen deficiency, counts how many rings plus pi bonds a molecule contains and is found from the formula $DoU = (2C + 2 - H)/2$ for a compound of carbon and hydrogen only. Substituting $C = 6$ and $H = 10$ gives $DoU = (2 	imes 6 + 2 - 10)/2 = (12 + 2 - 10)/2 = 4/2 = 2$. So the molecule has two degrees of unsaturation, which could be two double bonds, one triple bond, one ring plus one double bond, or two rings. A value of $1$ is wrong because it would correspond to formula $C_6H_{12}$, which has two more hydrogens. A value of $3$ is incorrect since it would require formula $C_6H_8$ with two fewer hydrogens. A value of $0$ would describe a fully saturated acyclic alkane $C_6H_{14}$, which does not match the given formula. This is the standard NCERT and JEE method of relating formula to unsaturation. A plausibility check: each missing pair of hydrogens relative to the saturated alkane corresponds to one degree of unsaturation, and $C_6H_{14}$ minus $C_6H_{10}$ is four hydrogens, giving two.

Degree Of Unsaturation

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